Limits / Series question 2

**champrock** · April 3rd 2009, 12:38 AM

Let a1 = 1 and an = n(an-1 + 1 ) for n=2,3,.....

Define

P = (1+ 1/a1 )(1+1/a2)......(1+1/an )

The Limit P when n tends to infinity is?

**Opalg** · April 3rd 2009, 04:47 AM

Originally Posted by champrock

Let a1 = 1 and an = n(an-1 + 1 ) for n=2,3,.....

Define

P = (1+ 1/a1 )(1+1/a2)......(1+1/an )

The Limit P when n tends to infinity is?

Step 1. Work out the first few terms to get an idea of what is happening. You should find that the first few values of $a_n$ are 1, 4, 15, 64, 325, 1956, and that the corresponding value of P is $\tfrac21.\tfrac54.\tfrac{16}{15}.\tfrac{65}{64}.\t frac{326}{325} .\tfrac{1957}{1956}$ . This simplifies to $\frac{1957}{6!} \approx2.718$ , which looks suspiciously close to e.

Now see if you can turn those observations into proofs. You need to show first that the product $(1+ 1/a_1 )(1+1/a_2)\ldots(1+1/a_n )$ is equal to $\frac{a_n+1}{n!}$ . Then you want to show that $\lim_{n\to\infty}\frac{a_n}{n!} = e$ .

Hint for that last part: let $b_n = \frac{a_n}{n!}$ . Show that $b_n = b_{n-1}+\frac1{(n-1)!}$ , and deduce that $b_n = 1+\frac1{1!}+\frac1{2!}+\frac1{3!}+\ldots+\frac1{( n-1)!}$ .

**champrock** · April 3rd 2009, 05:45 AM

answer is given as (e+1). ..

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